Quotient ring is a field

Question

I am asked whether $\mathbb{F}_4[x]/(x^2+x+1)$ is a field, where $\mathbb{F}_4$ is the field of $4$ elements.

I'm not too sure where to go with this. Since $\mathbb{F}_4$ is a field, it is a PID, and that polynomial is an ideal. However I'm not able to determine whether it is maximal or not, as I'm not sure how $\mathbb{F}_4$ looks like since it's not $\mathbb{Z}_4$ obviously. If there's some type of eisenstein's criteron for irreducibility in this field or anything I'd greatly appreciate some help.

Answer 1

$x^2+x+1$ is reducible with roots $\alpha$ and $\beta$ where $\alpha$ and $\beta$ are the elements of $\mathbb{F}_4$ not in $\mathbb{F}_2$.

We can also note every polynomial of degree 2 with coefficients in $\mathbb{F}_2$ has a root in a degree 2 extension of $\mathbb{F}_2$, therefore has a root in $\mathbb{F}_4$ which is the only degree 2 extension of $\mathbb{F}_2$.

Therefore $\mathbb{F}_4[x]/(x^2+x+1)$ is not a field.

Answer 2

Additional information:

If $\Bbb F_4=\{0,1,\omega,\omega+1\}$, where $\omega$ and $\omega+1$ are (primitive) cube roots of unity, both of them being roots of $X^2+X+1$, then we get, over $\Bbb F_4$, $$X^2+X+1=(X+\omega)(X+\omega+1)\,,$$ and it turns out that $\Bbb F_4[X]/(X^2+X+1)\cong\Bbb F_4\oplus\Bbb F_4$, where the isomorphism is induced by $X\mapsto(\omega,\omega+1)\in\Bbb F_4\oplus\Bbb F_4$.